Thread: Another nested material implication question

1. Another nested material implication question

I am interested in expressions of the form
((A → B) → C)

Specifically, given A, I would like to infer C.

I have found that
(((A → B) A) → (((A → B) → C) → C))

is a valid argument. But maybe I am headed off in the wrong direction. What I am trying to get to is a consequent of just C. Can I get there from here? How?

Thanks

2. Re: Another nested material implication question Originally Posted by anody I am interested in expressions of the for ((A → B) → C)
Specifically, given A, I would like to infer C.
It is not clear as to exactly what you mean, Is it this?
$\displaystyle \begin{array}{*{20}{c}} {(A \to B) \to C}\\ A\\ \hline {\therefore \quad C} \end{array}$ is a valid argument form.

3. Re: Another nested material implication question

Yes, except when I evaluate this in a truth table,

((((A → B) → C) ∧ A) → C)

it is not a tautology, hence not a valid argument.

A B C ((((A → B) → C) ∧ A) → C)
F F F T
F F T T
F T F T
F T T T
T F F F
T F T T
T T F T
T T T T

4. Re: Another nested material implication question Originally Posted by anody Yes, except when I evaluate this in a truth table,

((((A → B) → C) ∧ A) → C)

it is not a tautology, hence not a valid argument.

A B C ((((A → B) → C) ∧ A) → C)
F F F T
F F T T
F T F T
F T T T
T F F F
T F T T
T T F T
T T T T
You may think this is not valid BUT it is!
$\displaystyle \begin{array}{*{20}{c}} {(A \to B) \to C}\\ \neg A\\ \hline {\therefore \quad C} \end{array}$ SEE HERE

HERE is the proof:
$\displaystyle \begin{array}{*{20}{l}} {(A \to B) \to C}&\_&{\text{given}}\\ {\neg A}&\_&{\text{given}}\\ {\neg A \vee B}&\_&{\text{addition}}\\ {(A \to B)}&\_&{\text{Material Iplication}}\\ \hline {\therefore \;\quad C}&\_&{\text{Modus Ponens}} \end{array}$

5. Re: Another nested material implication question

Wow, beautiful! And no wonder I was unable to get from A to C.

...An afterthought:

((((A → B) → C) ∧ (A → (A → B))) → C)